Consider the following periodic Sindel sequence \begin{equation} 1\:2\:3\:4\:3\:2\:\:\:1\:2\:3\:4\:3\:2\:\:\:1\:2\: 3\:4\:3\:2 \end{equation}
The period sequence $1\:2\:3\:4\:3\:2$ is a sequence of numbers whose sum is the modulus $m$ and the period length is the number of terms in this sequence ($6$ in this exemple).
A Sindel sequence has the following nice summation property: it is possible to insert, in the periodic sequence, commas (or any other symbols) and $+$ signs so that the « visible sums » separated by the commas form the sequence of all the positive integers in their normal order. In our example, we have \begin{equation} 1,\: 2,\: 3,\: 4,\: 3+2,\: 1+2+3,\: 4+3,\: 2+1+2+3,\: 4+3+2,\: 1+2+3+4,\:... \end{equation} For every modulus $m$, there is a Sindel sequence. For example, for $m=17$, there is a Sindel sequence whose period sequence is $1\:1\:1\:1\:2\:4\:1\:4\:2$ and for $m=19$, the period sequence is $1\:1\:1\:3\:1\:2\:1\:5\:2\:2$. There are nice arithmetic problems related to those sequences, most of them using triangular numbers. A math professor asked the following:
Is it possible to choose a large enough $m$ in such a way that the number $2021$ appears in the corresponding Sindel sequence ?
Do you know if there is a way to find a solution using triangular numbers ? Many Thanks for any help.